Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $16.5$ years; the standard deviation is $2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living longer than $10.5$ years.
Solution: $16.5$ $14.5$ $18.5$ $12.5$ $20.5$ $10.5$ $22.5$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $16.5$ years. We know the standard deviation is $2$ years, so one standard deviation below the mean is $14.5$ years and one standard deviation above the mean is $18.5$ years. Two standard deviations below the mean is $12.5$ years and two standard deviations above the mean is $20.5$ years. Three standard deviations below the mean is $10.5$ years and three standard deviations above the mean is $22.5$ years. We are interested in the probability of a sloth living longer than $10.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the sloths will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the sloths will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $10.5$ years and the other half $({0.15\%})$ will live longer than $22.5$ years. The probability of a particular sloth living longer than $10.5$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.